Ferenc Tamas Gyurcsan wrote:
>
> >206.84.220.64/255.255.255.224.
> >It was explained to me that the net mask must be a bitstring of contiguous
> >'1's followed by a bitstring of contiguous '0's (starting at the MSB).
> >Using that rule, there's no way to construct an address/mask for the range
> >.65-.94.
>
> I'm still not sure about this, although I haven't tried it yet... Could you
> please give it a try and tell us about the result? I am really wondering about
> this because I read somewhere that it is possible.
That is correct. The `one' bits in a subnet mask MUST be contiguous.
Therefore there is a very small set of subnet masks (within a `C' class
address space), and therefore a small set of different sizes of subnets
(measured in hosts per subnet) that one can have.
255.255.255.0 00000000 254 hosts, plus network and broadcast
.128 10000000 126 hosts, " " " "
.192 11000000 62 hosts, " " " "
.224 11100000 30 hosts, " " " "
.240 11110000 14 hosts, " " " "
.248 11111000 6 hosts, " " " "
.252 11111100 2 hosts, " " " "
Notice that the number of hosts per subnet is always a power of 2 minus
2. I haven't figured out a trick for remembering the subnet mask's last
octet, but they're easy enough to memorize with so few of them.
You can not have a subnet mask of .254, because this leaves you with
only two possible combinations in each range, the network address and
the broadcast address... no room for hosts. And obviously a subnet mask
of .255 is (almost) completely useless...
--
PGP/GPG Public key at http://cerberus.ne.mediaone.net/~derek/pubkey.txt
Derek D. Martin | Senior UNIX Systems/Network Administrator
Arris Interactive | A Nortel Company
[EMAIL PROTECTED] | [EMAIL PROTECTED]
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